NAME
DATE
9-7
PERIOD
Study Guide and Intervention
Solving Linear-Nonlinear Systems
Systems of Equations Like systems of linear equations, systems of linear-nonlinear
equations can be solved by substitution and elimination. If the graphs are a conic section
and a line, the system will have 0, 1, or 2 solutions. If the graphs are two conic sections, the
system will have 0, 1, 2, 3, or 4 solutions.
Solve the system of equations. y = x 2 - 2x - 15
x + y = -3
Rewrite the second equation as y = -x - 3 and substitute it into the first equation.
-x - 3 = x2 - 2x - 15
0 = x2 - x - 12
Add x + 3 to each side.
0 = (x - 4)(x + 3)
Factor.
Use the Zero Product property to get
x = 4 or x = -3.
Substitute these values for x in x + y = -3:
4 + y = -3 or -3 + y = -3
y = -7
y=0
The solutions are (4, -7) and (-3, 0).
Exercises
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Solve each system of equations.
1. y = x2 - 5
y=x-3
2. x2 + ( y - 5)2 = 25
y = -x2
(2, -1), (-1, -4)
(0, 0)
3. x2 + ( y - 5)2 = 25
y = x2
4. x2 + y2 = 9
x2 + y = 3
(0, 3), ( √
5 , -2), (- √
5 , -2)
(0, 0), (3, 9), (-3, 9)
5. x2 - y2 = 1
x2 + y2 = 16
6. y = x - 3
x = y2 - 4
√
√
√
√
, − ), (− , - − ),
(−
√
√
√
√
, − ), (- − , - − )
(- −
34
2
34
2
Chapter 9
30
2
30
2
34
2
√
√
, − ),
(−
√
√
, −)
(−
30
2
34
2
29
7+
2
30
2
2
45
2
29
7-
29
1+
29
1-
2
Glencoe Algebra 2
Lesson 9-7
Example
NAME
DATE
9-7
PERIOD
Study Guide and Intervention
(continued)
Solving Linear-Nonlinear Systems
Systems of Inequalities
Systems of linear-nonlinear inequalities can be solved
by graphing.
Example 1
2
Solve the system of inequalities by graphing.
y
2
x + y ≤ 25
2
(x - −25 )
4
25
+ y2 ≥ −
2
4
The graph of x2 + y2 ≤ 25 consists of all points on or inside
the circle with center (0, 0) and radius 5. The graph of
-4
-2
2
4
x
2
4
x
-2
2
)
(
O
5
25
x-−
+ y2 ≥ −
consists of all points on or outside the
2
4
5
5
circle with center −
, 0 and radius −
. The solution of the
2
2
-4
( )
system is the set of points in both regions.
Example 2
2
Solve the system of inequalities by graphing.
y
2
x + y ≤ 25
4
y2
x2
−-−
>1
4
9
2
2
2
The graph of x + y ≤ 25 consists of all points on or inside
the circle with center (0, 0) and radius 5. The graph of
-4
-2
9
-4
the hyperbola shown. The solution of the system is the set of
points in both regions.
Exercises
Solve each system of inequalities by graphing.
y2
x2
+− ≤1
1. −
16
2. x2 + y2 ≤ 169
4
3. y ≥ (x - 2)2
x2 + 9y2 ≥ 225
1
y>−
x-2
(x + 1)2 + ( y + 1)2 ≤ 16
2
y
-4
Chapter 9
-2
y
y
4
12
4
2
6
2
O
2
4
x
-12 -6
O
6
12
x
-4
-2
O
-2
-6
-2
-4
-12
-4
46
2
4
x
Glencoe Algebra 2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
y
x2
−-−
> 1 are the points “inside” but not on the branches of
4
O
-2
2